MENSURATION

Mensuration of Heights and Distances.

By the application of geometry the measurement of lines, which, on account of their position or other circumstances, are inaccessible, is reduced to the determination of angles, and of other lines which are accessible, and admit of being measured by methods sufficiently obvious.

A line considered as traced on the ground may be measured with rods or a Gunter's chain of sixty-six feet; but more expeditiously with measuring tapes of fifty or one hundred feet. By these, if the ground be tolerably even, and the direction of the line be traced pretty correctly, a distance may, by using proper care, be measured within about three inches of the truth in every fifty feet, so that the error may not exceed the 200th part of the whole line.

Vertical angles may be measured with a quadrant furnished with a plummet and sights in the manner indicated by the annexed figures. If an angle of elevation is to be measured, as the angle contained by a horizontal line AC, (fig. 1.) and a line drawn from A to B the top of a tower, hill, or other eminence; or to a celestial body, as a star, &c.; the centre of the quadrant must be fixed at A, and the instrument moved about A, in the vertical plane, till to an eye placed at G, the object B be seen through the two sights D, d. Then will the arch EF, cut off by the plumb-line AF, be the measure of the angle CAB.

An angle of depression CAB (fig. 2), is to be measured exactly in the same manner, except that here the eye is to be placed at A the centre of the instrument, and the measure of the angle is the arch EF.

But the most convenient instrument for measuring angles, whether vertical or horizontal, is the Theodolite. This instrument has undergone various alterations and improvements since it was described (we believe, for the first time) by Sisson. Its principal parts are,

1. The horizontal limb. This consists of two circular plates AA, the upper or vernier plate, and BB, the graduated limb. The former moves freely above the latter, without actual contact, and both have a horizontal motion about a vertical axis C, which consists of two parts, one external, fixed to the graduated limb B, and another internal, fixed to the vernier plate A. Their form is conical, and they are ground and accurately fitted into each other, so as to have a perfectly steady motion. The external centre also fits into a ball at D, and the parts are held together by a screw at the lower end of the internal axis.

The lower plate is broader than the upper, and its edge is chamfered off and covered with silver to receive the graduations.

2. The verniers. These are short scales on the upper plate, and on opposite sides of it, or 180° asunder. They are chamfered, so as to form with the chamfered edge of the lower plate a continued conical surface, and are covered with silver, and graduated in such a manner as to sub-

divide the divisions of the lower plate into minutes. By Right lines means of microscopes which are generally fixed over the and angles. verniers, the half or even the fourth of a minute may be estimated.

3. The parallel plates F and G. These serve for levelling the instrument, and are held together by a ball and socket at D. Four milled-headed screws H, H, &c., enter and work in the lower plate, while their upper ends press against the lower side of the upper plate; and being set in pairs opposite to each other, they act in contrary directions, and by this means the instrument is set up level for observation.

Beneath the parallel plates is a female screw adapted to the staff-head, which is united by brass joints to three mahogany legs, so constructed, that when shut up they form one round staff, and are secured in that form by rings put on them. When opened out they make a very firm stand, and the parallel plates may be made horizontal, although the ground may be uneven.

The lower plate of the horizontal limb can be fixed in any position by tightening the clamping screw I, which causes the collar K to embrace the axis C, and prevents its moving; but the final adjustment is to be given by means of the slow-motion tangent screw L which is attached to the upper parallel plate. In like manner the upper or vernier plate can be fixed to the lower by a clamp M, which is also furnished with a slow motion, produced by a tangent screw.

4. Spirit levels N, N. There are two of these at right angles to each other, with proper adjusting screws, on the plane of the vernier plate. Their use is to determine when the horizontal limb is set truly level.

5. The compass. This is on the middle of the vernier plate, but of course it cannot be shewn in a side view of the instrument.

6. The vertical arc and telescope. The arc is placed on a horizontal axis, the ends of which are supported by two frames, (only one of which can be shewn in the figure.) At one end of the axis is an arm R, with a clamping screw S, by which the axis is held when the telescope is nearly in the proper position; and there is a tangent screw T, by which a slow angular motion is given to the vertical arc and telescope to direct the latter accurately to an object. One side of the vertical arc is inlaid with silver, and by means of its vernier V can be read off to single minutes. The other side shows the difference between the hypotenuse and the base of a right angled triangle, or the number of links to be deducted from each chain length to reduce hypotenusal to horizontal lines.

The level which is shewn under, and parallel to, the telescope is attached to it at one end by a joint, and at the other by a capstan headed screw for raising or lowering that end; there is another screw at the jointed end for lateral adjustment: by their united action the level is placed parallel to the optical axis or line of collimation of the telescope.

The telescope has two collars or rings of bell metal, ground truly cylindrical, on which it rests on its supports X, X, called Y's; and it is confined in its place by the clips Z, Z, which may be opened by removing the pins Y, Y, for the purpose of reversing the telescope or allowing it a circular motion round the axis during the adjustment.

In the focus of the eye glass are placed three lines formed of spider's web, one horizontal and two crossing it obliquely at equal angles so as to include a small angle between them. These are preferable to a single perpendicular wire, because a station staff at a distance can be made to bisect the angle formed by the oblique wires with more certainty than the staff can be bisected by a vertical wire.

Three of the four screws for adjusting the cross-wires are shewn at a, a, a, two of which are placed at right angles to the other two, so that by easing one of each pair and tightening the other opposite to it, the intersection of the wires may be placed exactly in the line of collimation.

Right lines and angles. The object glass of the telescope is fixed in a tube which slides within the outer tube, and may be thrust out or drawn backward by turning the milled head $b$ on the side of the telescope. By this motion the optical image of any object to which the telescope is directed may be brought to the intersection of the cross wires.

Before the theodolite can be applied to accurate practice its parts must be adapted to each other by means of the clamps and screws and levels. The operations by which this is performed are called adjustments.

The first adjustment is that of the line of collimation, which must coincide with the axis of the cylindrical rings on the telescope, and pass through the intersection of the cross wires. The second places the level attached to the telescope parallel to the rectified line of collimation. The third makes the axis of the horizontal limb truly vertical by means of the telescope level, which is most to be depended on; and then the levels on the vernier plate are adjusted by their screws so that their air bubbles may remain at rest in the middle of their tubes while it makes a complete revolution on its axis. When these adjustments are perfect the vernier of the vertical arc must be set so that its index may point to zero on the arc; or else its deviation from zero must be noted and applied as an index error.

PROBLEM I.

To determine the height of an object, the bottom of which is accessible.

Example. Having measured $AE$, a distance of 200 feet in a direct horizontal line from the bottom of a tower, the angle $BCD$, contained by the horizontal line $CD$; and a line drawn from $C$ to the top of the tower, was measured by a quadrant, or theodolite placed at $C$, and found to be $47^\{\backslash circ\}\; 30\text{'}$. The centre $C$ of the instrument was five feet above the line $AE$ at its extremity $E$. It is required hence to determine $AB$ the height of the tower.

In the right-angled triangle $CBD$ we have given the side $CD=200$ feet, and the angle $C=47^\{\backslash circ\}\; 30\text{'}$. And since by the rules of PLANE TRIGONOMETRY,

By employing logarithmic tables and proceeding as is taught in PLANE TRIGONOMETRY, we shall find $DB=218.3$ feet. To which add $DA=EC=5$ feet, the height of the instrument, and we have $AB=223.3$ feet, the height of the tower.

PROBLEM II.

To determine the height and distance of an object when the bottom or point directly under it is inaccessible.

Ex. 1. Suppose a small cloud, or balloon $C$, is seen at the same time by two observers at $A$ and $B$, and that these stations are in the same vertical plane with the object $C$, and on the same side of it. Also suppose that its angles of elevation, viz. the angles $A$ and $B$ are $35^\{\backslash circ\}$ and $64^\{\backslash circ\}$, and that $AB$, the distance between the observers, is 880 feet. It is required hence to determine $CD$ the height of the object, also $AC$, $BC$ its distances from the two observers.

In the triangle $CAB$ there are given the outward angle $CBD=64^\{\backslash circ\}$, and one of the inward angles $A=35^\{\backslash circ\}$; hence the other inward angle $ACB$, which is their difference, is given, and $=64^\{\backslash circ\}\; -\; 35^\{\backslash circ\}\; =\; 29^\{\backslash circ\}$.

Now in the triangle $CAB$

From these proportions, by actual calculation, $BC$ will be found $=1041$ feet, and $AC=1631$ feet.

Again, in the right-angled triangle $BCD$

Hence $CD$ will be found $=936$ feet.

Ex. 2. To measure the height of an obelisk $CD$, standing on the top of a declivity, two stations at $A$ and $B$ were

taken, one at the distance of forty, and the other at the distance of one hundred feet from the centre of its base, which was in a straight line with the stations. At the nearer station $A$, a line drawn from it to the top of the obelisk was found to make an angle of $41^\{\backslash circ\}$ with the plane of the declivity; and at $B$, the more remote station, the like angle was found to be $23^\{\backslash circ\}\; 45\text{'}$. Hence it is required to find the height of the obelisk.

From the angle $CAD=41^\{\backslash circ\}$, subtract the angle $B=23^\{\backslash circ\}\; 45\text{'}$, and there remains the angle $BCA=17^\{\backslash circ\}\; 15\text{'}$.

In the triangle $BCA$,
$\backslash text\{Sin.\; \}\; BCA\; :\; \backslash sin.\; B\; ::\; AB\; :\; AC$. Hence $AC=81.49$ feet.

And in the triangle $ACD$,
$AC+AD\; :\; AC-AD\; ::\; \backslash tan.\; \backslash frac\{1\}\{2\}(D+C)\; :\; \backslash tan.\; \backslash frac\{1\}\{2\}(D-C)$.
Hence $\backslash frac\{1\}\{2\}(D-C)=40^\{\backslash circ\}\; 24\backslash frac\{1\}\{2\}\text{'}$, which, subtracted from $\backslash frac\{1\}\{2\}(D+C)$ gives the angle $ACD=27^\{\backslash circ\}\; 5\backslash frac\{1\}\{2\}\text{'}$.

Lastly, in the triangle $ACD$,
$\backslash text\{Sin.\; \}\; ACD\; :\; \backslash sin.\; A\; ::\; AD\; :\; DC$.
Hence $DC$, the height required, will be found to be $57.62$ feet.

Ex. 3. At $B$ the top of a tower which stood on a hill near the sea shore, the angle of depression of a ship at anchor (viz. the angle $HBS$), was $4^\{\backslash circ\}\; 52\text{'}$; and at $R$, the bottom of the tower, its depression (namely, the angle $NRS$) was $4^\{\backslash circ\}\; 2\text{'}$. Required $AS$ the horizontal distance of the vessel; and also $RA$, the height of the bottom of the tower above the level of the sea, supposing $RB$ the height of the tower itself to be $54$ feet.

From the angle $BSA=HBS=4^\{\backslash circ\}\; 52\text{'}$, subtract the angle $RSA=NRS=4^\{\backslash circ\}\; 2\text{'}$, and there remains the angle $BRS=50^\{\backslash circ\}$. Also, from the angle $HBA=90^\{\backslash circ\}$ subtract $HBS=4^\{\backslash circ\}\; 52\text{'}$, and there remains $SBR=85^\{\backslash circ\}\; 8\text{'}$.

In the triangle $SBR$,
$\backslash text\{Sin.\; \}\; BSR\; :\; \backslash sin.\; SBR\; ::\; BR\; :\; SR$.
Hence $SR$ is found. Again, in the triangle $SRA$,
$\backslash text\{rad.\}\; :\; \backslash sin.\; RSA\; ::\; SR\; :\; AR$,
and $\backslash text\{rad.\}\; :\; \backslash cos.\; RSA\; ::\; SR\; :\; AS$.

From the first of these proportions we find $AR=260$ feet; and from the second, $AS=3690$ feet.

PROBLEM III.

To find the distance of an object from either of two given stations: also from a line joining them, when each station and the object can be seen from the other station.

Ex. 1. Wanting to know the breadth $CD$ of a river, and also the distance of an object $A$ close by its side from another object $C$ on its opposite side, a base $AB$ of 400 yards was measured along the bank. Then, by means of a theodolite, the angles $CBA$ and $CAB$ were measured, and found to be $37^\{\backslash circ\}\; 40\text{'}$ and $59^\{\backslash circ\}\; 15\text{'}$ respectively. It is required thence to determine the breadth $CD$, and the distance $AC$ between the objects $A$ and $C$.

This problem differs from the last only by the given angles, and distances required, lying in a horizontal instead of a vertical plane.

In the triangle $ABC$ we have the base $AB$, also the angles $A$ and $B$, and consequently the angle $C$ given. And by Plane Trigonometry,

Hence $AC$ is found to be $246.2$ yards.

Also, in the right-angled triangle $ACD$,
$\backslash text\{rad.\}\; :\; \backslash sin.\; A\; ::\; AC\; :\; CD$.

Hence $CD$ is found to be $211.6$ yards.

Ex. 2. From a ship at sea a point of land was observed to bear $E$. by $S$.; and after sailing $N.\; E.$ twelve miles, the same point was found to bear $S.\; E.$ by $E$. How far was the last observation made from the point of land?

Let $A$ be the first position of the ship, $B$ the second, and $C$ the point of land. In the triangle $ABC$ we have given the angle $A=5$ points or $56^\{\backslash circ\}\; 15\text{'}$, the angle $B=9$ points, or

Right lines $101^\{\backslash circ\}\; 15\text{'}$, and the angle $C=2$ points or $22^\{\backslash circ\}\; 30\text{'}$. Also the side $AB=12$ miles. Hence (by TRIGON.) the side $BC$ is readily found to be $26.073$ miles.

To determine the distance between two inaccessible objects, which can be seen from two given stations at a known distance from each other, supposing them all in the same plane.

Fig. 8. Ex. Wanting to know the distance between two inaccessible objects $H$ and $M$, a base $AB$ of $670$ yards was measured in the same plane with the objects, and the following angles were taken at its extremities.

Hence it is required to determine $HM$, the distance between the objects.

In the triangle $HAB$ we have the angle $HBA=42^\{\backslash circ\}\; 22\text{'}$, the angle $HAB\; (=HAM+MAB)=97^\{\backslash circ\}\; 56\text{'}$, and therefore the remaining angle $AHB=39^\{\backslash circ\}\; 42\text{'}$. We have also the side $AB=670$ yards. Hence, by this proportion,

we find $AH=706.8$ yards.

Again, in the triangle $MAB$ we have the angle $MAB=40^\{\backslash circ\}\; 16\text{'}$, the angle $ABM\; (=ABH+HBM)=113^\{\backslash circ\}\; 29\text{'}$, and therefore the angle $AMB=26^\{\backslash circ\}\; 15\text{'}$. Hence from the proportion,

we get $AM=1389.4$.

In the triangle $HAM$, besides the angle $HAM=57^\{\backslash circ\}\; 40\text{'}$ we have now the sides $AH=706.8$, and $AM=1389.4$ yards, to find the remaining side $HM$. Therefore, proceeding according to the rules of trigonometry, we state this proportion,

Hence we find half the difference of the angles $AHM$ and $AMH$ to be $30^\{\backslash circ\}\; 36\text{'}$, which taken from $61^\{\backslash circ\}\; 10\text{'}$, half the sum, leaves $30^\{\backslash circ\}\; 34\text{'}$ for $AMH$ the least of the two angles. Lastly, from the proportion

we get $HM=1174$ yards, the answer to the question.

The mutual distances of three remote objects being given, and the angles which these distances subtend at a station, in their plane, to find the position of the station.

Example. There are three objects, $A,\; B,\; C$, whose distances asunder are known to be as follows: namely, from $A$ to $B$ $106\backslash frac\{1\}\{2\}$, from $A$ to $C$ $202$, and from $B$ to $C$ $131$ fathoms. Now to determine the distance of $D$ a fourth object, or station, from each of the other three, the angle $ADB$ was measured with a theodolite, or other suitable instrument, and found to be $13^\{\backslash circ\}\; 30\text{'}$, and the angle $CDB$ was found $29^\{\backslash circ\}\; 50\text{'}$. Hence it is required to determine the distances $DA,\; DB$, and $DC$, supposing $DB$ the least of the three.

Let a circle be described through the points $A,\; D$, and $C$; and let $DB$ be produced to meet the circle again in $E$, and draw $AE,\; CE$.

In the triangle $AEC$ there are given the side $AC=202$ fathoms, the angle $ACE\; (=ADE)$. GEOM. Sect. II. Theor.

$15^\{\backslash circ\}\; 30\text{'}$, and the angle $CAE\; (=CDE)$ $29^\{\backslash circ\}\; 50\text{'}$. Hence (by TRIGON.) we shall have $AE=68716$ fathoms.

In the triangle $ABC$, all its sides are given, and hence the angle $BAC$ will be found $=35^\{\backslash circ\}\; 35\text{'}\; 54\text{'}\text{'}$; to this, add the angle $CAE$, and the sum is the angle $EAB=65^\{\backslash circ\}\; 25\text{'}\; 54\text{'}\text{'}$.

In the triangle $ABE$, we have given $AB=106.5$, $AE=68716$, the angle $BAE=65^\{\backslash circ\}\; 25\text{'}\; 54\text{'}\text{'}$; hence we shall have the angle $ABE=38^\{\backslash circ\}\; 43\text{'}\; 41\text{'}\text{'}$, and the angle $AEB=75^\{\backslash circ\}\; 51\text{'}\; 25\text{'}\text{'}$.

In the triangle $ADE$ we have the side $AE=68716$, the angle $ADE=13^\{\backslash circ\}\; 30\text{'}$, and the angle $AED=75^\{\backslash circ\}\; 51\text{'}\; 25\text{'}\text{'}$. Hence we have $AD=285.43$ fathoms, which is one of the distances required.

In the triangle $ABD$ we have $AB=106.5$, the angle $ADB=13^\{\backslash circ\}\; 30\text{'}$, the angle $DAB\; (=ABE-ADB)$ $25^\{\backslash circ\}\; 13\text{'}\; 45\text{'}\text{'}$. Hence $BD$, another of the distances sought, will be found $=194.45$ fathoms.

Lastly, in the triangle $ADC$, there is given $AC=202$, the angle $ADC\; (=ADB+BDC)$ $=43^\{\backslash circ\}\; 20\text{'}$, the angle $DCA\; (=DEA)=75^\{\backslash circ\}\; 15\text{'}\; 25\text{'}\text{'}$. Hence we get $DC=256.97$ fathoms, which is the remaining distance sought.

There are various other instruments and methods by which the heights or distances of objects may be found. One of the most simple instruments, both in respect of its construction and application, is a square, $ABCD$, made of some solid material, and furnished with two sights on $AB$, one of its edges, and a plummet fastened to $A$, one of its angles, and having the two sides $BC,\; CD$, which contain the opposite angle divided into 10, or 100, or 1000 equal parts.

To measure any altitude $HK$ with this instrument. Let it be held in such a position that

$K$, the top of the object may be seen through the sights on its edge $AB$, while its plane is perpendicular to the horizon; then the plummet will cut off from the square a triangle similar to that formed by the horizontal line $AI$, the vertical line $IK$, and the line $AK$ drawn from the eye to the top of the object.

If the line of the plummet pass through $D$ the opposite angle of the square, then the height $KI$ will be equal to $AI$, the distance of the eye from the vertical line to be measured. If it meet $AD$, the side of the square next the eye, in some point $E$ between $A$ and $D$, then the triangles $ABE,\; ABE$, being similar, and the angle $ABE$ equal to the angle $AKI$, we have $AE\; :\; AB\; ::\; AI\; :\; IK$. Let us now suppose $AD=AB$ to be divided into 1000 equal parts; then the length of $AE$ will be expressed by a certain number of these parts; thus the proportion of $AE$ to $AB$, and consequently that of $AI$ to $IK$ will be given; therefore if $AI$ be determined by actual measurement, we may from the above proportion immediately find $IK$.

If again the line of the plummet meet $DC$ the side of the

Plane figures. square opposite to the sights, in F, then, in the similar triangles AIK, BCF, the angle AKI is equal to BFC; thus we have BC : CF :: AI : IK. Hence IK is determined as before, and in each case by adding HI the height of the eye, we shall have HK the whole height required.

To find the area of a parallelogram, whether it be a square, a rectangle, a rhombus, or a rhomboid.

Multiply the length by the perpendicular breadth, and the product will be the area.

This rule is demonstrated in GEOMETRY, sect. iv. theor. 5.

Ex. 1. Required the area of a square ABCD, whose side AB is $10\backslash frac\{1\}\{2\}$ inches.

Here $10\backslash frac\{1\}\{2\}\; \backslash times\; 10\backslash frac\{1\}\{2\}$ or $10.5\; \backslash times\; 10.5\; =\; 110.25$ square inches is the area required.

Ex. 2. Required the area of a rectangle EFGH, whose length EF is 13.75 chains, and breadth FG is 9.5 chains.

Here $13.75\; \backslash times\; 9.5\; =\; 130.625$ square chains is the area, which, when reduced to acres, &c. is 13 ac. 0 ro. 10 po.

Ex. 3. Required the area of a parallelogram KLMN, whose length KL is 37 feet, and perpendicular breadth NO is $5\backslash frac\{1\}\{2\}$ or 5.25 feet.

In this example the area is $37\; \backslash times\; 5.25\; =\; 194.25$ square feet, or 21.583 square yards.

As radius,

To the sine of any angle of the parallelogram,
So is the product of the sides including the angle,
To the area of the parallelogram.

To see the reason of this rule it is only necessary to observe, that in the parallelogram KLMN, the perpendicular breadth NO is a fourth proportional to radius, sine of the angle K, and the oblique line KN, (TRIGONOMETRY,) and

is therefore equal to $\backslash frac\{\backslash sin.\; K\}\{\backslash text\{rad.\}\}\; \backslash times\; \backslash text\{KN\}$; therefore the area

of the figure is $\backslash frac\{\backslash sin.\; K\}\{\backslash text\{rad.\}\}\; \backslash times\; \backslash text\{KN\}\; \backslash times\; \backslash text\{KL\}$, which expression is the same as the result obtained by the above rule.

Ex. Suppose the sides KL and KN are 36 feet, and 25.5 feet, and the angle K is $58^\backslash circ$, required the area.

Here it will be convenient to employ a table of logarithms. The operation may stand thus,

Having given any two sides of a right angled triangle, to find the remaining side.

1. When the sides about the right angle are given, to find the hypotenuse.

Add together the squares of the sides about the right angle, and the square root of the sum will be the hypotenuse.

2. When the hypotenuse and one of the sides about the right angle is given, to find the other side.

From the square of the hypotenuse subtract the square of the given side, and the square root of the remainder will be the other side.

This rule is deduced from theor. 13. sect. iv. GEOMETRY.

Ex. 1. In a right-angled triangle ABC, the sides AB and AC, about the right angle, are 33 feet and 56 feet; what is the length of the hypotenuse BC?

Here $33^2\; +\; 56^2\; =\; 3136\; +\; 1089\; =\; 4225$, and $\backslash sqrt\{4225\}\; =\; 65$ feet, = the hypotenuse BC.

Ex. 2. Suppose the hypotenuse BC to be 65 feet, and AB one of the sides about the right angle to be 33 feet; what is the length of AC the other side?

Here $65^2\; -\; 33^2\; =\; 4225\; -\; 1089\; =\; 3136$; and $\backslash sqrt\{3136\}\; =\; 56$ feet = the side AC.

To find the area of a triangle.

Multiply any one of its sides by the perpendicular let fall upon it from the opposite angle, and half the product will be the area.

The truth of this rule is proved in GEOMETRY, Part I. sect. iv. theor. 6.

Example. What is the area of a triangle ABC, whose base AC is 40, and perpendicular BD is 14.52 chains?

The product of the base by the perpendicular, or $40\; \backslash times\; 14.52$, is 580.8 square chains, the half of which, or 290.4 sq. ch. = 29 ac. 0 r. 6.4 po. is the area of the triangle.

As radius,

To the sine of any angle of a triangle,
So is the product of the sides including the angle,
To twice the area of the triangle.

This rule follows immediately from the second rule of Prob. I., by considering that the triangle KNL is half the parallelogram KNML.

Example. What is the area of a triangle ABC, whose two sides AB and AC are 30 and 40, and the included angle A is $28^\backslash circ\; 57\text{'}$?

When the three sides are given, add together the three sides, and take half the sum. Next, subtract each side severally from the said half sum, thus obtaining three remain-

Plane figures. Lasts. Lastly, multiply the said half sum, and those three remainders all together, and extract the square root of the last product for the area of the triangle.

This practical rule is deduced from the following geometrical theorem. The area of a triangle is a mean proportional between two rectangles, one of which is contained by half the perimeter of the triangle, and the excess of half the perimeter above any one of its sides; and the other is contained by the excesses of half the perimeter above each of the other two sides. As this theorem is not only remarkable, but also of great utility in mensuration, we shall here give its demonstration.

Let ABC then be any triangle; produce AB, any one of its sides, and take BD and Bd, each equal to BC; join CD and Cd, and through A draw a line parallel to BC, meeting CD and Cd produced in E and e; thus the angle AED will be equal to the angle BCD, (GEOMETRY, Part I. sect. i. theor. 21), that is, to the angle BDC or ADC, (sect. i. theor. 11); and hence AE = AD, (sect. i. theor. 12); and in like manner, because the angle Aed is equal to the angle Bcd, that is, to the angle Bdc, or Ade, therefore Ae = Ad.

On A as a centre, at the distance AD or AE, describe a circle meeting AC in F and G; and on the same centre, with the distance Ad or Ae, describe another circle meeting AC in f and g, and draw BH and Bh perpendicular to CD and Cd. Then, because BD, BC, Bd are equal, the point C is in the circumference of a circle, of which Dd is the diameter, therefore CD and Cd are bisected at H and h, (sect. ii. theor. 6), and the angle Dcd is a right angle, (sect. ii. theor. 17), and hence the figure CHBh is a rectangle, so that Bh = CH = 1/2 CD, and Bh = Ch = 1/2 Cd.

Join BE, and Be, then the triangle BAC is equal to each of the triangles BEC, BeC, (sect. iv. theor. 2, cor. 2); but the triangle BEC is equal to 1/2 EC x BH, (sect. iv. theor. 2), that is, to 1/2 EC x Cd; and in like manner the triangle BeC is equal to 1/2 eC x Bh, that is, to 1/2 eC x Cd, therefore the triangle ABC is equal to 1/2 EC x Cd, and also to 1/2 eC x Cd.

Now since CD : Cd :: CE x CD : CE x Cd } sect. iv. }
and also CD : Cd :: Ce x CD : Ce x Cd } theor. 3. }

Therefore CE x CD : CE x Cd :: Ce x CD : Ce x Cd; that is, because CE x CD = FC x CG, and Ce x Cd = fC x Cg, (sect. iv. corollaries to theor. 28 and 29).

$FC\; \backslash times\; CG\; :\; CE\; \backslash times\; Cd\; ::\; Ce\; \backslash times\; CD\; :\; fC\; \backslash times\; Cg$; which last proportion (by taking one-fourth of each of its terms, and substituting the triangle ABC for its equivalent values 1/2 CE x Cd and 1/2 Ce x CD) gives us
$\backslash frac\{1\}\{2\}\; FC\; \backslash times\; \backslash frac\{1\}\{2\}\; CG\; :\; \backslash text\{trian.\; ABC\}\; ::\; \backslash text\{trian.\; ABC\}\; :\; \backslash frac\{1\}\{2\}\; fC\; \backslash times\; \backslash frac\{1\}\{2\}\; Cg$.

Now, if it be considered that the radius of the circle gde is AB = BC, it will readily appear that, putting 2s for the perimeter of the triangle ABC, we have

Put now a, b, c for the sides AC, BC, and AB respectively, then 1/2 FC = s, 1/2 GC = s - a, 1/2 fC = s - b, 1/2 Cg = s - c; thus the last proportion becomes

$s\; \backslash times\; (s\; -\; a)\; :\; \backslash text\{trian.\; ABC\}\; ::\; \backslash text\{trian.\; ABC\}\; :\; (s\; -\; b)\; \backslash times\; (s\; -\; c)$, which conclusion, when expressed in words at length, is evidently the proposition to be demonstrated.

And as a mean proportional between two quantities is found by taking the square root of the product, it follows that the area of the triangle ABC, which is a mean between $s\; \backslash times\; (s\; -\; a)$ and $(s\; -\; b)\; \backslash times\; (s\; -\; c)$, is equal to

which formula, when expressed in words at length, gives the preceding rule.

Example. Required the area of a triangle whose three sides are 24, 36, and 48 chains respectively.

Here $24\; +\; 36\; +\; 48\; =\; 108\; =$ the sum of the three sides,

And $\backslash frac\{108\}\{2\}\; =\; 54\; =$ half that sum;

Also $54\; -\; 24\; =\; 30$, the first remainder; $54\; -\; 36\; =\; 18$, the second remainder; and $54\; -\; 48\; =\; 6$, the third remainder.

The product of the half sum and remainders is

And the square root of this product is

PROBLEM IV.

To find the area of a trapezoid.

RULE.

Add together the two parallel sides, then multiply their sum by the perpendicular breadth, or distance between them, and half the product will be the area.

This rule is demonstrated in GEOMETRY, Part I. sect. iv. theor. 7.

Example. Required the area of the trapezoid ABCD, whose parallel sides AB and DC are 7.5 and 12.25 chains and perpendicular breadth DE is 15.4 chains.

The sum of the parallel sides is $7.5\; +\; 12.25\; =\; 19.75$; which multiplied by the breadth, is

and half this product is

the area required.

PROBLEM V.

To find the area of any trapezium.

RULE.

Divide the trapezium into two triangles by a diagonal, then find the areas of these triangles, and add them together.

Note. If two perpendiculars be let fall on either diagonal from the other two opposite angles, the sum of these per-

pendiculars being multiplied by the diagonal, half the product will be the area of the trapezium. The reason of this rule is sufficiently obvious.

Example. In the trapezium ABCD the diagonal AC is 42, and the two perpendiculars BE, DF are 16 and 18: What is its area?

Here the sum of the perp. is $16+18=34$, which multiplied by 42, and divided by 2, gives

To find the area of an irregular polygon.

Draw diagonals dividing the proposed polygon into trapeziums and triangles; then find the areas of all these separately, and add them together for the content of the whole polygon. The reason of this rule, and the manner of applying it, are sufficiently obvious.

To find the area of a regular polygon.

Multiply the perimeter of the polygon, or sum of its sides, by the perpendicular drawn from its centre on one of its sides, and take half the product for the area.

This rule is only in effect resolving the polygon into as many triangles as it has sides, by drawing lines from its centre to all its angles, then taking the sum of their areas for the area of the figure.

Example. Required the area of a regular pentagon ABCDE, whose side AB or BC, &c., is 25 feet, and perpendicular HK is 17.2 feet.

Here $25\; \backslash times\; 5\; =\; 125\; =$ the perimeter,

And $125\; =\; 17.2\; =\; 2160$,

And its half $1075\; =$ the area required.

Note. If only the side of the polygon be given, its perpendicular may be found by the following proportion:

As radius,

To the tan. of half the angle of the polygon,

So is half the side of the polygon,

To the perpendicular.

And here, as well as in all other trigonometrical calculations, we may employ a table of logarithmic sines and tangents.

The angle of the polygon, that is, the angle contained by any two of its adjacent sides, will be found from this theorem. The sum of all its interior angles is equal to twice as many right angles, wanting four, as it has sides, which is demonstrated in GEOMETRY, Part I. sect. i. theor. 25.

To find the diameter and circumference of a circle, the one from the other.

As 7 is to 22, so is the diameter to the circumference, nearly.

As 22 is to 7, so is the circumference to the diameter, nearly.

As 113 is to 355, so is the diameter to the circumference, nearly.

As 355 is to 113, so is the circumference to the diameter, nearly.

As 1 is to 3.1416, so is the diameter to the circumference, nearly.

As 3.1416 is to 1, so is the circumference to the diameter, nearly.

Note. The result obtained by the first rule, which is the least accurate of the three, will not differ from the true answer by so much as its 2400th part. But that obtained by the second rule, which is the most accurate, will not differ by so much as its 10,000,000th part.

The proportion of the diameter of a circle to its circumference, is investigated in ALGEBRA, art. 273, and in GEOMETRY, Part I. sect. vi. prop. 6. Also in FLUXIONS, sect. 38 and sect. 139. The manner of finding the first and second rules, and others of the same kind, is explained in ALGEBRA, sect. xxi. But it is impossible to express exactly, by finite numbers, the proportion of the diameter of the circle to its circumference.

Example 1. To find the circumference of a circle whose diameter is 20.

By the first rule,

Or by the third rule, $3.1416\; \backslash times\; 20\; =\; 62.832$ the answer.

Ex. 2. The circumference of a circle is 10 feet, what is its diameter?

By the second rule,

To find the length of any arch of a circle.

As 180 is to the number of degrees in the arch, so is 3.1416 times the radius to its length.

To see the reason of this rule it is only necessary to consider, that 3.1416 times the radius is (by last rule) equal to half the circumference, or to an arch of $180^\backslash circ$, and that the length of an arch is proportional to the number of degrees it contains.

Example. Required the length of the arch AEB, whose chord AB is 6, the radius AC or CB being 9. Draw CD perpendicular to the chord, then CD will bisect the chord in D, and the arch in E. Now in the right-angled triangle ACD, there is given the hypotenuse AC=9, and the side AD=3; hence, by trigonometry, the angle ACE will be found to contain $19^\backslash circ\; 28\text{'}\; \backslash frac\{1\}\{2\}\; =\; 19.471$ degrees. The double of this, or 38.942, is the number of degrees in the whole arch AEB. Then, by the rule,

the answer.

As six times the distance of any chord from the centre together with nine times the radius of the circle, is to that distance together with fourteen times the radius; so is the chord to the arc nearly.

The investigation of this rule is as follows:

Let $a$ denote any arc of a circle of which the radius = 1. Putting $A,\; B,\; C$ for indeterminate quantities, assume

Then, observing that $\backslash sin.\; a\; \backslash cos.\; a\; =\; \backslash frac\{1\}\{2\}\; \backslash sin.\; 2a$, we have $\backslash sin.\; a\; +\; \backslash frac\{1\}\{2\}\; A\; \backslash sin.\; 2a\; =\; a\; B\; +\; a\; C\; \backslash cos.\; a$. (1.)

Now, by formulae investigated in ALGEBRA, art. 286,

Therefore we have for $\backslash sin.\; a,\; \backslash frac\{1\}\{2\}\; \backslash sin.\; 2a,\; a\; \backslash cos.\; a$; these approximate values, which include the fifth power of the arc,

Hence substituting these values in equation (1.) and dividing by $a$ we obtain

Making now the coefficients of like terms on each side of this equation equal, we have

These values being substituted for $A,\; B,\; C$ in the assumed equation, the result after reduction is

Hence the rule.

Ex. 1. Required the length of $\backslash frac\{1\}\{6\}$th of the circumference of a circle whose radius is 10?

In this case the chord $AB$ is equal to the radius $AC=10$; and the perpendicular $CD=\backslash sqrt\{(CA^2-AB^2)\}=$

$\backslash sqrt\{(100-25)\}=\backslash sqrt\{75\}=8.660254$; hence $6\; CD+9\; AC=141.961524$, and $CD+14\; AC=148.660254$. We now state this proportion,

This is a near approximation to the length of the arc; its true value is 10.471976, &c.

Ex. 2. It is known that an arc of $57^\backslash circ\; 17\text{'}\; 44\text{'}\text{'}\; .8$ is very nearly equal to the radius. Now, let it be proposed to find its approximate length by the formula, the radius being supposed equal to 1 mile.

In this case by the Trigonometrical Tables the chord = 958851; its distance from the centre = 8775826. Therefore the proportion is

The error of the approximation is .0000078 of a mile, which is equal to about half an inch.

To find a straight line nearly equal to a given arc of a circle.

The formula $\backslash frac\{14+\backslash cos.\; a\}\{9+6\; \backslash cos.\; a\}\; =\; \backslash frac\{a\}\{\backslash sin.\; a\}$ may be put under this form

Hence we obtain the following simple geometrical construction for finding a straight line nearly equal to a given arc of a circle:

In $DL$ produced take $LH$ equal to the radius, and $HK$ from $H$ towards the centre equal to one fifth part of the versed sine $DE$: From $K$ draw $KF,\; KG$ through $A$ and $B$ the extremities of the arc $AB$ to meet the tangent at the midpoint of the arc in $F$ and $G$: The intercepted portion $FG$ of the tangent $GH$ is equal to the arc $AB$ nearly.

To find the area of a circle.

Multiply half the circumference by half the diameter, and the product will be the area.

Multiply the square of the diameter by .7854, and the product will be the area.

The first of these rules has been demonstrated in GEOMETRY, sect. vi. prop. 3. And the second rule is deduced from the first, as follows. It appears from prop. 6. sect. vi. GEOMETRY, that the diameter of a circle being unity, its circumference is 3.1416 nearly; therefore, by the first rule, its area is $1\; \backslash times\; 3.1416\; \backslash div\; 4\; =\; .7854$. But circles are to one another as the squares of their diameters, (prop. 4.) therefore, putting $d$ for the diameter of any circle, $1\; :\; d^2\; ::\; .7854\; :\; .7854\; d^2\; =$ the area of the circle whose diameter is $d$.

Example. What is the area of a circle whose diameter is 7.

By the second rule $7\; \backslash times\; 7\; \backslash times\; .7854\; =\; 38.4846$ the area.

By the first rule $7\; \backslash times\; 3.1416\; =$ the circumference.

Then $\backslash frac\{7\; \backslash times\; 3.1416\; \backslash times\; 7\}\{4\}\; =\; 7\; \backslash times\; 7\; \backslash times\; .7854\; =\; 38.4846$ the area, the same as before.

To find the area of any sector of a circle.

Multiply the radius by half the arch of the sector, and the product will be the area, as in the whole circle.

As 360 is to the degrees in the arc of the sector, so is the area of the whole circle to the area of the sector.

The first of these rules follows easily from the rule for the whole area, by considering that the whole circumference is to the arch of the sector, as the whole area to the area of the sector, that is,

Example. To find the area of a circular sector ACB whose arch AEB contains 18 degrees, the diameter being 3 feet. (See fig. to prob. ix.)

Find the area of the sector having the same arch with the segment by the last problem. Find also the area contained by the chord of the segment and the two radii of the sector. Then take the sum of these two for the answer when the segment is greater than a semicircle, or take their difference when it is less than a semicircle. As is evident by inspection of the figure of a segment. (Fig. to prob. ix.)

Example. To find the area of the segment AEBDA, its chord AB being 12, and the radius AC or BC 10.

First, as $AC\; :\; AD\; ::\; \backslash text\{rad.\}\; :\; \backslash sin.\; 36^\backslash circ\; 52\text{'}\; \backslash frac\{1\}\{2\}\; =\; 36\; \backslash cdot\; 87$ degrees, the degrees in the angle ACE or arch AE. And their double or $73\; \backslash cdot\; 74\; =$ the degrees in the whole arch AEB.

Now $7854\; \backslash times\; 400\; =\; 31416$ the area of the whole circle. Therefore $360^\backslash circ\; :\; 73\; \backslash cdot\; 74\; ::\; 31416\; :\; 643504\; =$ area of the sector CAEB.

Again $\backslash sqrt\{(CA^2\; -\; AD^2)\}\; =\; \backslash sqrt\{(100\; -\; 36)\}\; =\; \backslash sqrt\{64\}\; =\; 8\; =\; DC$.

Therefore $AD\; \backslash times\; DC\; =\; 6\; \backslash times\; 8\; =\; 48\; =$ area of the triangle. Hence sector ACBA - triangle ACB = $16\; \backslash cdot\; 3504$ the area of seg. AEBDA.

To find the area of any segment of a parabola, that is the space included by any arch of a parabola, and the straight line joining its extremities.

Multiply the base of the segment by its height, and take $\backslash frac\{2\}\{3\}$ of the product for the area.

This rule is demonstrated in CONIC SECTIONS, part iv. sect. iii. prop. 2.

Example. The base AB of a parabolic segment ACB is 10, and its altitude CD, (that is, the greatest line that can be drawn in the segment perpendicular to the base AB) is 4: What is its area?

Multiply the product of the two axes by the number 7854 for the area of the ellipse.

For the area of an ellipse is equal to the area of a circle whose diameter is a mean proportional between the axes of the ellipse, (as is easily deduced from CONIC SECTIONS, cor. i. prop. 3. sect. iii. part iv.) that is, to the area of a circle, the square of whose diameter is equal to the product of the axes. But by prob. x. the

area of a circle is equal to the square of the diameter multiplied by 7854; therefore the area of an ellipse is equal to the product of the axes multiplied by the same number 7854.

Example. If the axes of an ellipse, ABCD, be 35 and 25. What is the area?
$35\; \backslash times\; 25\; \backslash times\; 7854\; =\; 687225$ the area.

Note. As to hyperbolic areas, the reader will find formulas for their exact mensuration in FLUXIONS, § 156. Ex. 4 and 5.

To find nearly the area of a figure bounded by any curve line A a a' a'', &c. P, the straight line BQ, and AB, PQ two other straight lines drawn from the extremities of the curve perpendicular to BQ.

Let BQ, the base of the figure, be divided into any even number of equal parts by the perpendiculars ba, b'a', b''a'', &c. which meet the curve in the points a, a', a'', &c.

Let F and L denote the first and last perpendiculars AB and PQ.

Let E denote the sum of all the remaining even perpendiculars, viz. a b, a'' b'', a'' b'', the second, fourth, sixth, &c.

Let R denote the sum of the remaining perpendiculars, viz. a' b', a''' b''', &c.

And put D for B b, or b b', &c. the common distance between the perpendiculars.

Then the area of the figure will be nearly equal to $\backslash frac\{1\}\{2\}\; D\; \backslash times\; (F\; +\; L\; +\; 4E\; +\; 2R)$; and the approximation will be so much the more accurate according as the number of perpendiculars is the greater.

Demonstration. Join the tops of the first and third perpendiculars by the line A a' meeting the second perpendicular in E, and draw CD through a so as to form the parallelogram A a' DC; then the space bounded by the curve line A a a' and the three straight lines AB, B b', b' a' will be made up of the trapezoid AB b' a', and the space bounded by the arch A a a' and its chord A a'. Now if the arch A a a' be small, this last space will be nearly two thirds of the parallelogram AD, for it will be nearly equal to the area contained by the straight line A a', and an arch of a parabola passing through the points A, a, a', and having a b for a diameter, which area is $\backslash frac\{2\}\{3\}$ of its circumscribing parallelogram. (CONIC SECTIONS, part iv. sect. iii. prop. 2.) Therefore the space A a a' b' BA will be nearly equal to the sum of the trapezoid AB b' a' and $\backslash frac\{2\}\{3\}$ of the parallelogram AD, which sum is evidently equal to $\backslash frac\{1\}\{2\}$ of the trapezoid AB b' a', together with $\backslash frac\{2\}\{3\}$ of the trapezoid CB b' D. Now the area

of the trapezoid AB b' a' is $\backslash frac\{AB\; +\; a\text{'}b\text{'}\}\{2\}\; \backslash times\; Bb\text{'}$ (GEOMETRY,

sect. iv. theor. 7.) $=\; \backslash frac\{AB\; +\; a\text{'}b\text{'}\}\{2\}\; \backslash times\; 2Bb$; and in like man-

ner the area of the trapezoid CB b' D is $\backslash frac\{CB\; +\; Dv\text{'}\}\{2\}\; \backslash times\; Bb\text{'}\; =$

$ab\; \backslash times\; 2Bb$; therefore the area of the figure A a a' b' B is nearly $\backslash frac\{1\}\{2\}\; \backslash times\; \backslash frac\{AB\; +\; a\text{'}b\text{'}\}\{2\}\; \backslash times\; 2Bb\; +\; \backslash frac\{2\}\{3\}\; \backslash times\; ab\; \backslash times\; 2Bb\; =\; \backslash frac\{1\}\{2\}\; (AB\; +\; 4ab\; +\; a\text{'}b\text{'})\; Bb$.

Plane figures. In the very same way it may be shewn that the area of the figure $a\text{'}\; a\text{'}\text{'}\; a\text{'}\text{'}\text{'}\; b\text{'}\text{'}\; b\text{'}$ is nearly $\backslash frac\{1\}\{2\}(a\text{'}\; b\text{'}\; +\; a\text{'}\text{'}\; b\text{'}\text{'}\; +\; a\text{'}\text{'}\text{'}\; b\text{'}\text{'}\text{'})\; \backslash times\; Bb$, and that the area of the figure $a\text{'}\text{'}\; a\text{'}\text{'}\text{'}\; a^\{iv\}\; PQ\; b\text{'}\text{'}\text{'}$ is nearly $\backslash frac\{1\}\{2\}(a\text{'}\text{'}\; b\text{'}\text{'}\text{'}\; +\; a\text{'}\text{'}\text{'}\; b^\{iv\}\; +\; PQ)\; \backslash times\; Bb$.

Therefore, the area of the whole figure bounded by the curve line AP, and the straight lines AB, BQ, QP, is nearly equal to the sum of these three expressions, namely, to

as was to be demonstrated.

Example 1. Let it be required to find the area of the quadrant ABC, whereof the radius AC=1.

Let AC be bisected by the perpendicular DE, and let CD be divided into four equal parts by the perpendiculars $mn,\; pq,\; rs$. Now because CA=1, therefore CD=$\backslash frac\{1\}\{2\}$, Cr=$\backslash frac\{1\}\{4\}$, Cp=$\backslash frac\{1\}\{8\}$, Cm=$\backslash frac\{1\}\{16\}$. Hence DE=$\backslash sqrt\{(EC^2\; -\; CD^2)\}\; =\; \backslash sqrt\{(1\; -\; \backslash frac\{1\}\{4\})\}\; =\; \backslash frac\{1\}\{2\}\backslash sqrt\{3\}$; and in like manner $rs\; =\; \backslash frac\{1\}\{8\}\backslash sqrt\{55\},\; pq\; =\; \backslash frac\{1\}\{4\}\backslash sqrt\{15\},\; mn\; =\; \backslash frac\{1\}\{2\}\backslash sqrt\{63\}$. Therefore,

Ex. 2. To find the area of the hyperbola FDM, of which the absciss FM=10, the semiordinate MD=12, and semitransverse CF=15.

Let FM be divided into five equal parts by the semiordinates HI, $mn,\; pq,\; rs$. Thus CH=17, Cm=19, Cp=21, Cr=23, CM=25.

Now, since from the nature of the curve, $\backslash sqrt\{(CM^2\; -\; CF^2)\}\; :\; MD\; ::\; \backslash sqrt\{(CH^2\; -\; CF^2)\}\; :\; HI$ (CONIC SECTIONS, part iii. prop. 22. and GEOM. part i. sect. iv. theor. 12.) that is, in numbers, $20\; :\; 12\; ::\; 8\; :\; HI$, therefore $HI\; =\; \backslash frac\{3\}\{2\}$. In like manner we find $mn\; =\; \backslash frac\{1\}\{2\}\backslash sqrt\{34\},\; pq\; =\; \backslash frac\{1\}\{2\}\backslash sqrt\{6\},\; rs\; =\; \backslash frac\{1\}\{2\}\backslash sqrt\{19\}$. Therefore

The figure HIDM = $103.2762\; \backslash times\; \backslash frac\{1\}\{2\}\; =\; 68.8508$, to which adding FIH, considered as a portion of a parabola, we have 75.245 for the area of the hyperbola.

OF LAND-SURVEYING.

The instruments most commonly employed in land-surveying, are the Chain, the Plane Table, and Cross.

A statute acre of land being 160 square poles, the chain is made four poles, or sixty-six feet in length, that ten square chains, or 100,000 square links) may be equal to an acre. Hence each link is 7.92 inches in length.

The plane table is used for drawing a plan of a field, and taking such angles as are necessary to calculate its area. It is of a rectangular form, and is surrounded by a moveable frame, by means of which a sheet of paper may be fixed to its surface. It is furnished with an index by which a line

may be drawn on the paper in the direction of any object in the field, and with scales of equal parts by which such lines may be made proportional to the distances of the objects from the plane table when measured by the chain, and its frame is divided into degrees for observing angles.

The cross consists of two pair of sights set at right angles to each other upon a staff having a pike at the bottom to stick into the ground. Its use is to determine the points where a perpendicular drawn from any object to a line will meet that line; and this is effected by finding by trials a point in the line, such that the cross being fixed over it so that one pair of the sights may be in the direction of the line, the object from which the perpendicular is to be drawn may be seen through the other pair; then the point thus found will be the bottom of the perpendicular, as is evident.

A theodolite may also be applied with great advantage to land-surveying, more especially when the ground to be measured is of great extent.

In addition to these, there are other instruments employed in surveying, as the perambulator, which is used for measuring roads and other great distances. Levels, with telescopic or other sights, which are used to determine how much one place is higher or lower than another. An offset-staff for measuring the offsets and other short distances. Ten small arrows, or rods of iron or wood, which are used to mark the end of every chain length. Pickets or staves with flags to be set up as marks or objects of direction; and, lastly, scales, compasses, &c. for protracting and measuring the plan upon paper.

The observations and measurements are to be regularly entered as they are taken, in a book which is called the field book, and which serves as a register of all that is done or occurs in the course of the survey.

To Measure a Field by the Chain.

Let AmBCnDq represent a field to be measured. Let it be resolved into the triangles AmB, ABD, BCD, AqD. Let all the sides of the large triangles ABD, BCD, and the perpendiculars of the small ones AmB, AqD from their vertices m, q, be measured by the chain, and the areas calculated by the rules delivered in this section, and their amount is the area of the whole. But if, on account of the curvature of its sides the field cannot be wholly resolved into triangles, then, either a straight line may be drawn over the curve side, so that the parts cut off from the field, and those added to it may be nearly equal; or, without going beyond the bounds of the field, the curvilinear spaces may be measured by the rule given in prob. xvi. of this section.

To Measure a Field with the Plane Table.

Let the plane table be fixed at F, about the middle of the field ABCDE, and its distances FA, FB, FC, &c. from the several corners of the field measured by the chain. Let the index be directed from any point assumed on the paper to the points A, B, C, D, &c., successively, and the lines Fa, Fb, Fc, &c., drawn in these directions.

Let the angles contained by these lines be observed, and the lines themselves made proportional to the distances measured. Then their extremities being joined, there will be formed

Solids. a figure $a\; b\; c\; d\; e$ similar to that of the field; and the area of the field may be found by calculating the areas of the several triangles of which it consists.

To Plan a Field from a given Base Line.

Let two stations $A,\; B$ be taken within the field, but not in the same straight line with any of its corners; and let their distance be measured. Then the plane table being fixed at $A$, and the point $a$ assumed on its surface directly above $A$, let its index be directed to $B$, and the straight line $a\; b$ drawn along the side of it to represent $AB$. Also, let the index be directed from $a$ to an object at the corner $C$, and an indefinite straight line drawn in that direction, and so of every other corner successively. Next, let the plane table be set at $B$, so that $b$ may be directly over $B$, and $b\; a$ in the same direction with $BA$, and let a straight line be drawn from $b$ in the direction $BC$. The intersection of this line with the former, it is evident will determine the point $C$, and the triangle $a\; b\; c$ on the paper will be similar to $ABC$ in the field. In this manner all the other points are to be determined, and these being joined there will be an exact representation of the field.

If the angles at both stations were observed, as the distance between them is given, the area of the field might be calculated from these data, but the operation is too tedious for practice. It is usual therefore to measure such lines in the figure that has been constructed as will render the calculation easy.

SECTION III.—MENSURATION OF SOLIDS.

PROBLEM I.

To find the surface of a right prism, or cylinder.

RULE.

Multiply the perimeter of the end by the length or height of the solid, and the product will be the surface of all its sides; to which add also the area of the two ends of the prism when required.

The truth of this rule will be evident, if it be considered that the sides of a right prism are rectangles, whose common length is the same as the length of the solid, and their breadths taken altogether make up the perimeter of the ends of the prism. And as a cylinder may be considered as the limit of all the prisms which can be inscribed in or circumscribed about its base, so the surface of the cylinder will be the limit of the surfaces of these prisms, and the expression for that limit is evidently the product of the circular base by its height. Or a cylinder may be considered as a prism of an indefinitely great number of sides.

Ex. 1. What is the surface of a cube, the length of its side $AB$ being 20 feet?

Here $4\; \backslash times\; 20\; =\; 80$ the perim. of end.

And $80\; \backslash times\; 20\; =\; 1600$ the four sides.

And $2\; \backslash times\; 20\; \backslash times\; 20\; =\; 800$ the top and bottom.

The sum $2400\; =$ the area or surface.

Ex. 2. What is the convex surface of a cylinder whose length $AB$ is 20 feet, and the circumference of its base 3 feet? Here $3\; \backslash times\; 20\; =\; 60$ feet, the answer.

PROBLEM II.

To find the surface of a right pyramid or cone.

RULE.

Multiply the perimeter of the base by the slant height or length of the side, and half the product will evidently be the surface of the sides, or the sum of the areas of all the triangles which form it. To which add the area of the end or base, if required.

Note. Here a cone is considered as a pyramid of an indefinitely great number of sides.

Ex. 1. What is the upright surface of a triangular pyramid, $ABCD$, the slant height $AE$, being 20 feet, and each side of the base 3 feet?

Here $\backslash frac\{3\; \backslash times\; 3\; \backslash times\; 20\}\{2\}\; =\; 90$ feet, the surface.

Ex. 2. Required the convex surface of a cone, the slant height $AB$ being 50 feet, and the diameter of its base $8\backslash frac\{1\}\{2\}$ feet.

Here $8\backslash frac\{1\}\{2\}\; \backslash times\; 3\backslash frac\{1\}\{2\}\; \backslash times\; 50\; =\; 6667\backslash frac\{1\}\{2\}$ feet.

And $\backslash frac\{8\backslash frac\{1\}\{2\}\; \backslash times\; 3\backslash frac\{1\}\{2\}\; \backslash times\; 50\}\{2\}\; =\; 6667\backslash frac\{1\}\{2\}$, the answer.

PROBLEM III.

To find the surface of the frustum of a right pyramid or cone, being the lower part, when the top is cut off by a plane parallel to the base.

RULE.

Add together the perimeters of the two ends, and multiply their sum by the slant height, and take half the product for the answer.

The truth of this rule will be evident if it be considered that the sides of the frustum are trapezoids, whose parallel sides bound its top and base, and whose common breadth is its slant height.

Example. How many square feet are in the surface of a frustum $AG$ of a square pyramid, (see fig. to prob. vi.) whose slant height $RE$ is 10 feet; also each side of the greater end $AC$ is 3 feet 4 inches, and each side of the lesser end $EG$ 2 feet 2 inches?

Here $3\backslash frac\{1\}\{3\}\; \backslash times\; 4\; =\; 13\backslash frac\{1\}\{3\}$ the per. of gr. end.

And $2\backslash frac\{1\}\{4\}\; \backslash times\; 4\; =\; 8\backslash frac\{1\}\{2\}$ the per. of less end.

And their sum is 22 feet.

Therefore $\backslash frac\{22\; \backslash times\; 10\}\{2\}\; =\; 110$ feet, is the answer.

PROBLEM IV.

To find the solid content of any prism or cylinder.

RULE.

Find the area of the base or end of the figure, and multiply it by the height or length, and the product will be the area.

This rule follows immediately from theor. 11, and theor. 2, part ii. sect. iii. GEOMETRY.

Ex. 1. What is the solid content of a cube $AG$, the length of whose sides is 24 inches?

Here $24\; \backslash times\; 24\; =\; 576$ sq. inches, the area of the end.

And $576\; \backslash times\; 24\; =\; 13824$ cub. inches is the solidity.

Ex. 2. Required the content of a triangular prism, whose

Solids length AD is 20 feet, and the sides of its triangular base ABC, are 3, 4, and 5 feet.

First, the area of the triangular base is found by Rule iii. of Prob. 3, sect. ii. to be

Therefore $6\; \backslash times\; 20\; =\; 120$ cub. feet, the solidity.

Ex. 3. The Winchester bushel is a cylinder $18\backslash frac\{1\}\{2\}$ inches in diameter, and eight inches deep. How many cubic inches does it contain?

By prop. 10 of sect. ii. the area of its base is

Therefore $268803\; \backslash times\; 8\; =\; 2150424$ is the solid content.

PROBLEM V.

To find the solid content of any pyramid or cone.

RULE.

Find the area of the base, and multiply that area by the height, and one-third of the product will be the content of the solid.

This rule is demonstrated in GEOMETRY, part ii. sect. ii. prop. xvii. cor. 1, and sect. iii. theor. 3.

Ex. 1. What is the content of a triangular pyramid ABCD, whose perpendicular height AF is 30 feet, and each side of its base BCD is three feet? (See fig. in prob. ii. sect. iii.)

First, the area of the base, as found by rule iii. of prob. 3, sect. ii. is

Therefore $\backslash frac\{3.89711\; \backslash times\; 30\}\{3\}\; =\; 38.9711$ cub. feet is the solid content.

Ex. 2. What is the solid content of a cone, the radius BC of its base being nine inches, and its height AC 15 feet? (See fig. in prob. ii. sect. iii.)

Here $7854\; \backslash times\; \backslash frac\{3^2\}\{2\backslash pi\}\; =\; 1.76715$ is the area of the base in square feet.

And $\backslash frac\{1.76715\; \backslash times\; 15\}\{3\}\; =\; 8.8357$ cub. feet is the solid content.

PROBLEM VI.

To find the solidity of the frustum of a cone or pyramid.

RULE.

Add into one sum the areas of the two ends, and the mean proportional between them, that is, the square root of their product, and one-third of that sum will be a mean area, which being multiplied by the perpendicular height or length of the frustum will give the content.

Demonstration. Let PABCD be any pyramid, and AG a frustum of it contained between ABCD its base, and EFGH, a plane parallel to the base. Put $a$ for the side of a square equal to AC the base of the frustum; $b$ for the side of a square equal to EG its top; $h$ for LM the height of the frustum, and $c$ for PL the height of the part of the pyramid above the frustum. Then $a^2$ is the area of the base of the frustum; $b^2$ is the area of its top; $\backslash frac\{1\}\{3\}\; a^2\; (h+c)$ is the solid content of the whole pyramid, (GEOM. part ii. sect. ii. theor. 17), $\backslash frac\{1\}\{3\}\; b^2\; c$ is the content of its upper part; and therefore

is the solid content of the frustum itself. Now the base and top of the frustum being similar figures, (GEOM. part ii. sect. ii. theor. 13), their areas are to one another as the squares of AB and EF, their homologous sides, (part i. sect. iv. theor. 27). But $AB\; :\; EF\; ::\; BP\; :\; PF$ (sect. iv. theor. 20.) :: $PM\; :\; PL$; therefore the area of the base of the frustum is to the area of its top as $PM^2\; :\; PL^2$, that is, $a^2\; :\; b^2\; ::\; (h+c)^2\; :\; c^2$, and consequently $a\; :\; b\; ::\; h+c\; :\; c$;

hence $ac\; =\; b(h+c)$, and $c\; =\; \backslash frac\{bh\}\{a-b\}$, and $h+c\; =\; \backslash frac\{ah\}\{a-b\}$. Let these values of $c$ and $h+c$ be now substituted in the preceding expression for the content of the frustum, and it will become by proper reduction,

Let the numerator of the fractional part of this formula be actually divided by its denominator, and we shall obtain for the area of the frustum this more simple expression,

which formula, when expressed in words, is the rule. And as a cone may be considered as the limit of all the pyramids that can be inscribed in it, when the number of sides is conceived indefinitely increased, it is evident that the rule will apply alike to the cone and pyramid.

Ex. 1. Required the solidity of the frustum of a hexagonal pyramid, the side of whose greater end is four feet, and that of its lesser end is three feet, and its height nine feet.

First, by prob. 7, sect. ii. the area of the base of the frustum is found to be 41.569, and the area of its lesser end 23.383 square feet. And the mean proportional between these is

Hence the mean area is

And the solid content of the frustum is

Ex. 2. What is the solidity of the frustum of a cone, the diameter of the greater end being five feet, that of the lesser end three feet, and the altitude nine feet?

Here the area of the greater end is (by prob. 10, sect. ii.) $5^2\; \backslash times\; 7854$, and the area of the lesser end is $3^2\; \backslash times\; 7854$, and the mean proportional between them is $\backslash sqrt\{(5^2\; \backslash times\; 3^2\; \backslash times\; 7854)\}\; =\; 5\; \backslash times\; 3\; \backslash times\; 7854$; therefore the mean area is

And the content of the frustum

PROBLEM VII

To find the surface of a sphere, or of any segment or zone of it.

RULE.

Multiply the circumference of the sphere by the height of the part required, and the product will be the curve surface, whether it be a segment, a zone, or the whole sphere.

Note. The height of the whole sphere is its diameter. The truth of this rule has been already shown in the article FLUXIONS, sect. 165. It may, however,

be deduced from principles more elementary, by reasoning as follows: Let PCQ be a semi-circle, and AB, BC, CD, &c., several successive sides of a regular polygon inscribed in it. Con-

ceive the semicircle to revolve about the diameter PQ as an axis, then the arch ABCDE will generate a portion of the surface of a sphere, and the chords AB, BC, CD, &c., will generate the surfaces of frustums of cones; and it is easy to see that the number of chords may be so great that the surface which they generate shall differ from the surface generated by the arch ACE by a quantity which is less than any assigned quantity. Bisect AB in L, and draw AF, LM, BG, CH, &c., perpendicular to PQ. For the sake of brevity, let circ. AF denote the circumference of a circle whose radius is AF. Then because AF, BG, LM, are to each other respectively as circ. AF, circ. BG, circ. LM, (GEOM. part i. sect. vi. prop. 4. cor. i.), and because $\backslash frac\{1\}\{2\}(\backslash text\{AF\}\; +\; \backslash text\{BG\})\; =\; \backslash text\{LM\}$, therefore $\backslash frac\{1\}\{2\}(\backslash text\{circ.\; AF\}\; +\; \backslash text\{circ.\; BG\})\; =\; \backslash text\{circ.\; LM\}$. Now the area of the surface generated by the chord AB is $\backslash frac\{1\}\{2\}(\backslash text\{circ.\; AF\}\; +\; \backslash text\{circ.\; BG\})\; \backslash times\; \backslash text\{AB\}$, (prob. 3), therefore the same area is also equal to $(\backslash text\{circ.\; LM\})\; \backslash times\; \backslash text\{AB\}$. Draw AO parallel to FG, and draw LN to the centre of the circle. Then the triangles AOB, LMN are manifestly similar; therefore $\backslash text\{AB\}\; :\; \backslash text\{AO\}\; ::\; \backslash text\{NL\}\; :\; \backslash text\{LM\}\; ::\; \backslash text\{circ.\; NL\}\; :\; \backslash text\{circ.\; LM\}$; and hence $\backslash text\{AO\}\; \backslash times\; \backslash text\{circ.\; NL\}\; =\; \backslash text\{AB\}\; \backslash times\; \backslash text\{circ.\; LM\}$. But this last quantity has been proved equal to the surface generated by AB, therefore the same surface is equal to $\backslash text\{AO\}\; \backslash times\; \backslash text\{circ.\; NL\}$, or to $\backslash text\{FG\}\; \backslash times\; \backslash text\{circ.\; NL\}$, that is, to the rectangle contained by FG and the circumference of a circle inscribed in the polygon. In the same way it may be shown that the surfaces generated by BC, CD, DE, are respectively equal to $\backslash text\{GH\}\; \backslash times\; \backslash text\{circ.\; LN\}$, $\backslash text\{IK\}\; \backslash times\; \backslash text\{circ.\; LN\}$, $\backslash text\{JK\}\; \backslash times\; \backslash text\{circ.\; LN\}$. Therefore the whole surface generated by the chords AB, BC, CD, DE, &c., is equal to $(\backslash text\{FG\}\; +\; \backslash text\{GH\}\; +\; \backslash text\{HI\}\; +\; \backslash text\{IK\})\; \backslash times\; \backslash text\{circ.\; LN\}\; =\; \backslash text\{FK\}\; \backslash times\; \backslash text\{circ.\; LN\}$. Conceive now the number of chords between A and E to be indefinitely increased; then, observing that the limit of the surface generated by the chords is the surface generated by the arch ABCDE, and that the limit of NL is NP, the radius of the generating circle, it follows that the spherical surface or zone generated by the arch ACE is equal to the product of the circumference of the sphere by FK the height of the zone.

Ex. 1. What is the superficies of a globe whose diameter is 17 inches?

First $17\; \backslash times\; 3\; \backslash cdot\; 1416\; =\; 53\; \backslash cdot\; 4072$ inches = the circum.

Then $53\; \backslash cdot\; 4072\; \backslash times\; 17\; =\; 907\; \backslash cdot\; 9224$ sq. inches = 6305 sq. feet, the answer.

Ex. 2. What is the convex surface of a segment eight inches in height cut off from the same globe?

Here $53\; \backslash cdot\; 4072\; \backslash times\; 8\; =\; 427\; \backslash cdot\; 2576$ sq. inches = 2967 sq. feet, the answer.

To find the solidity of a sphere.

Multiply the area of a great circle of the sphere by its diameter, and take $\backslash frac\{2\}\{3\}$ of the product for the content.

Multiply the cube of the diameter by the decimal .5236 for the content.

The first of these rules is demonstrated in GEOMETRY, part i. sect. iii. theor. 6. And the second is deduced from the first, thus: put $d$ for the diameter of the sphere, then $d^2\; \backslash times\; 7854$ is the area of a great circle of the sphere, and by the first rule $\backslash frac\{2\}\{3\}\; d\; \backslash times\; d^2\; \backslash times\; 7854\; =\; d^3\; \backslash times\; 5236$ is its content.

Example. What is the content of a sphere whose diameter is six feet?

Answer $6^3\; \backslash times\; 5236\; =\; 113\; \backslash cdot\; 0976$ cub. feet.

To find the solid content of a spherical segment.

From three times the diameter of the sphere take double the height of the segment, then multiply the remainder by the square of the height, and the product by the decimal .5236 for the content.

This rule has been investigated in FLUXIONS, sect. 163. But it may be proved in a more elementary manner by means of the following axiom. If two solids be contained between two parallel planes, and if the sections of these solids by a third plane parallel to the other two, at any altitude, be always equal to one another, then the solids themselves are equal. Or more generally thus. If two solids between two parallel planes be such, that any sections of them by a third plane parallel to the other two have always to each other the same given ratio, then the solids themselves are to one another in that ratio. We have given this proposition in the form of an axiom, for the sake of brevity, but its truth may be strictly demonstrated, as has been done when treating of pyramids and the sphere, in GEOMETRY, part ii. sect. ii. and iii.

Let us now suppose ABE to be a quadrant, C the centre of the circle, AFEC a square described about the quadrant, and CF the diagonal of the square. Suppose the figures to revolve about AC as an axis, then the quadrant will generate a hemisphere, the triangle ACF will generate a cone, and the square AE a cylinder. Let these three solids be cut by a plane perpendicular to the axis, and meeting the plane of the square, in the line DHBG, and join CB. Then because CDB is a right-angled triangle, a circle described with CB as a radius is equal to two circles described with CD and DB as radii. (GEOM. part i. sect. vi. prop. 4, cor. 2.) But $\backslash text\{CB\}\; =\; \backslash text\{DG\}$, and since $\backslash text\{CA\}\; =\; \backslash text\{AF\}$, therefore $\backslash text\{CD\}\; =\; \backslash text\{DH\}$; therefore the circle described with the radius DG, is equal to the sum of the circles described with the radii DH, DB; that is, the section of the cylinder at any altitude, is equal to the corresponding sections of the sphere and cone taken together. Consequently, by the foregoing axiom, the cylinder is equal to the hemisphere and cone taken together, and also the segment of the cylinder between the planes AF, DG is equal to the sum of the segments of the hemisphere and cone contained between the same planes. Put 2 CE, or 2 AF, the diameter of the circle, $=\; d$, and AD, the height of the spherical segment, $=\; h$. Then $\backslash text\{AC\}\; =\; \backslash frac\{1\}\{2\}d$ and $2\backslash text\{CA\}\; =\; 2\backslash text\{AD\}\; =\; 2\backslash text\{CD\}\; =\; d\; -\; 2h$. Let $n$ denote the number 7854. Then the area of the base common to the conic frustum AH, and cylinder AG, is $n\; d^2$, (sect. ii. prob. 10), and the area of the top of the frustum is $n(d\; -\; 2h)^2$, and the mean proportional between these areas is $n(d\; -\; 2h)d$. Therefore the solid content of the frustum is (by prob. 6 of this sect.)

Now the solid content of the cylinder is $n\; d^2\; h$: (prob. 1). Therefore the solid content of the spherical segment, (which is equal to the difference between the cylinder AG and the conic frustum AH), is equal to

that is, to $2\; n\; d\; h^2\; -\; \backslash frac\{2\}\{3\}\; n\; h^3$, or to

which expression, if it be considered that $\backslash frac\{2\; n\}\{3\}$ or $\backslash frac\{2\; \backslash times\; 7854\}\{3\}$

Solids. is equal to .5236, is evidently the same as that given by the rule.

Example. In a sphere whose diameter is 21, what is the solidity of a segment whose height is 4.5 inches? First $3\; \backslash times\; 21\; =\; 2\; \backslash times\; 4.5\; =\; 54$. Then $54\; \backslash times\; 4.5\; \backslash times\; 4.5\; \backslash times\; .5236\; =\; 572.5566$ inches, the solidity required.

PROBLEM X.

To find the solid content of a paraboloid, or solid, produced by the rotation of a parabola about its axis.

RULE.

Multiply the area of the base by the height, and take half the product for the content.

To demonstrate this rule, let AGC and BHD be two equal semi-parabolas lying in contrary directions, and having their vertices at the extremity of the line AB. Let AD, BC be ordinates to the curves. Complete the rectangle ABCD, and conceive it to revolve about AB as an axis; then the rectangle will generate a cylinder, the radius of whose base will be AD, and the two semi-parabolas will generate two equal paraboloids having the same base and altitude as the cylinder. Let a plane be drawn perpendicular to the axis, and let FHGE be the common section of this plane and the generating figure. Let P denote the parameter of the axis. Then since

Hence it appears, as in the demonstration of the preceding rule, that of the solids described by ADCB, ACB, ADB between the same parallel planes, the section of the cylinder at any altitude is equal to the corresponding sections of the paraboloids taken together. Consequently (by the axiom) the cylinder is equal to both the paraboloids taken together; hence each is half a cylinder of the same base and altitude agreeing with the rule.

The same thing is also proved in FLUXIONS, sect. 163.

Example. If the diameter of the base of a paraboloid be 10 and its height 12 feet, what is its content? Here $10\; \backslash times\; 7.854\; =\; 7.854$ the area of the base. And $\backslash frac\{1\}\{2\}\; \backslash times\; 7.854\; \backslash times\; 12\; =\; 47.124$ cub. feet is the solidity.

PROBLEM XI.

To find the solid content of a frustum of a paraboloid.

RULE.

Add together the areas of the circular ends, then multiply that sum by the height of the frustum, and take half the product for its solid content.

To prove this rule put A and a for the greater and lesser ends of the frustum, and h for its height; also let c denote the height of the portion cut off from the complete paraboloid, so as to form the frustum. Then, by the last problem, the content of the complete paraboloid is $\backslash frac\{1\}\{2\}\; A(h+c)$, and the content of the part cut off is $\backslash frac\{1\}\{2\}\; ac$, therefore the content of the frustum is

But from the nature of the parabola, $c\; :\; h\; +\; c\; ::\; a\; :\; A$; hence $A\; c\; =\; a\; h\; +\; a\; c$ and $c\; =\; \backslash frac\{ah\}\{A-a\}$.

Let this value of c be substituted instead of it in the above expression for the content of the frustum, and it becomes

and hence is derived the rule.

Example. Required the solidity of the frustum of a paraboloid, the diameter of the greater end being 58, and that of the lesser end 30, and the height 18.

First, (by prob. 10, sect. ii.) we find the areas of the ends to be $58^2\; \backslash times\; 7.854$, and $30^2\; \backslash times\; 7.854$; therefore their sum is $(58^2\; +\; 30^2)\; \backslash times\; 7.854\; =\; 4264\; \backslash times\; 7.854$; and the content of the figure is $\backslash frac\{1\}\{2\}\; \backslash times\; 4264\; \backslash times\; 7.854\; \backslash times\; 18\; =\; 3140.5104$, the answer.

PROBLEM XI.

To find the solid content of a parabolic spindle or solid generated by the rotation of AEB an arc of a parabola, about AB an ordinate to the axis.

RULE.

Multiply the area of the middle section by the length, and take $\backslash frac\{1\}\{2\}$ of the product for the content of the solid.

For the investigation of this rule we must refer the reader to FLUXIONS, § 163. Ex. 4.

Example. The length of the parabolic spindle AEB is 60, and the middle diameter Ee 34; what is the solidity?

Here $34^2\; \backslash times\; 7.854$ is area of the middle section. Therefore $34^2\; \backslash times\; 7.854\; \backslash times\; 60\; \backslash times\; \backslash frac\{1\}\{2\}\; =\; 29053.5168$ is the solidity required.

PROBLEM XII.

To find the solid content of the frustum of a parabolic spindle, one of the ends of the frustum passing through the centre of the spindle.

RULE.

Add into one sum eight times the square of the diameter of the greater end, and three times the square of the lesser end, and four times the product of the diameters; multiply the sum by the length, and the product multiplied by .5236, or $\backslash frac\{1\}\{2\}$ of 7.854, will be the content.

For, referring the reader to FLUXIONS, § 163. Ex. 4. as before, and substituting $h$ for $AC\; =\; \backslash frac\{1\}\{2\}\; b$, but, in other respects, retaining the figure and notation, we have this general expression for the segment AP,

which, when $x=h$ gives $\backslash frac\{8\backslash pi\; h^5\}\{15a^2\}$ for the value of the semi-spindle. From this quantity let the former be subtracted, and there will remain

for the content of the frustum. In this expression let $z$ be put instead of $h-x$ or CD, and, denoting CE the radius of the greater end of the spindle by $d$, let $\backslash frac\{h^2\}\{d\}$ be substituted instead of its value $a$. Then we shall have the content of the frustum otherwise expressed by

which value, by putting $h\backslash sqrt\{\backslash frac\{d\}\{d\}\}$ in its two last terms instead of $z$, is changed to

and hence is derived the preceding rule.

Example. Suppose the diameter of the greater end to be 8, and the diameter of the lesser end 6, and the length 10, required the content?

To find the solid content of a spheroid, or solid generated by the rotation of an ellipse about either axis.

Multiply continually together the fixed axis, and the square of the revolving axis, and the number .5236 or $\backslash frac\{1\}\{6\}$ of 3.1416, and the last product will be the solidity.

For, let the semiellipse ADB, and semicircle AEB, revolve about the same fixed axis AB, and thus generate a spheroid and sphere. Let CD the revolving semiaxis of the ellipse meet the circle in E, and draw QP any ordinate to the fixed axis meeting the circle in R. Then, from the nature of the ellipse $PQ^2\; :\; PR^2$

$\backslash therefore\; CD^2\; :\; CE^2$ or $CA^2\; :\; CB^2$ (CONIC SECTIONS, part ii. prop. 13, cor. 3.) Hence it follows, (GEOM. part i. sect. vi. prop. 4.), that every section of the spheroid is to the corresponding section of the sphere in the same given ratio, namely, that of the square of the revolving axis to the square of the fixed axis; therefore (Axiom in the dem. of prob. 9.) the whole spheroid is to the whole sphere in the same ratio. That is, (because the content of the sphere is $AB^3\; \backslash times\; .5236$) $AB^3\; :\; (2CD^2)\; ::\; AB^3\; \backslash times\; .5236$: (the cont. of spheroid.) Hence the content of the spheroid is $AB\; \backslash times\; (2CD^2)\; \backslash times\; .5236$.

Ex. 1. What is the solid content of an oblong spheroid, or solid generated by the rotation of an ellipse about its greater axis, the axes being 50, and 30?

Ex. What is the solid content of an oblate spheroid, or solid generated by the rotation of an ellipse about its lesser axis, the two axes being as before?

To find the solid content of the frustum of a spheroid, its ends being perpendicular to the fixed axis, and one of them passing through the centre.

To the area of the less end add twice that of the greater, multiply the sum by the altitude of the frustum, and $\backslash frac\{1\}\{3\}$ of the product will be the content.

Note. This rule will also apply to the sphere.

Demonstration. Let ABE be a quadrant of an ellipse C its centre, CAFE its circumscribed rectangle, and CF its diagonal. Draw any straight line DG parallel to CE, meeting AC, CF, ABE and EF in D, H, B, and G. Then by CONIC SECTIONS, part ii. prop. 13.

and by sim. tr. $AF^2\; :\; DH^2\; ::\; CA^2\; :\; CD^2$.

Therefore (GEOMETRY, part i. sect. iii. theor. 10.),

Conceive now the figure to revolve about AC as an axis, so that the elliptic quadrant may generate the half of a spheroid, the rectangle AEF a cylinder, and the triangle ACF a cone; then it is evident (as was shewn in the case of the sphere in prob. 9.) that every section of the first of these solids by a plane perpendicular to the axis is equal to the difference of the sections of the other two, and consequently that the frustum of the spheroid between CE and DG is equal to the difference between the cylinder having DG or CE for the radius of its base, and CD for its altitude, and the cone having DH for the radius of its base, and DC for its altitude.

Put $n$ for the number 3.1416, then (prob. 4.) the content of the cylinder is $4n\; \backslash times\; DG^2\; \backslash times\; CD$, and (prob. 5.) the content of the cone is $\backslash frac\{1\}\{3\}n\; \backslash times\; DH^2\; \backslash times\; CD$, and therefore the content of the frustum of the spheroid is

But it was shewn that $DH^2\; =\; DG^2\; -\; DB^2$; therefore the content of the frustum is also $CE^2$ equal to

and hence is derived the rule.

Ex. Suppose the greater end of the frustum to be 15, the less end 9, and the length 10 inches, required the content?

The area of the greater end is $15^2\; \backslash times\; .7854$, and the area of the less end $9^2\; \backslash times\; .7854$, therefore the content is $.7854\; (9^2\; +\; 2\; \backslash times\; 15^2)\; \backslash times\; \backslash frac\{1\}\{3\}\; =\; 1390.158$ cubic inches.

To find the solid content of a hyperboloid, or solid generated by the rotation of a hyperbola about its transverse axis.

As the sum of the transverse axis and the height of the solid is to the sum of the said transverse axis and $\backslash frac\{2\}\{3\}$ of the height so is half the cylinder of the same base and altitude to the solidity of the hyperboloid.

Demonstration. Let $BAb$ be a hyperbola, $Aa$ its transverse axis, $C$ its centre, $CF,\; Cf$ its asymptotes, $Faf$ a tangent at its vertex. Draw $FE$ parallel to $CA$, and draw any straight line parallel to $Ff$, meeting the asymptotes in $H$ and $h$, the curve in $B$ and $b$, the axis in $D$, and the line $FE$ in $G$. Then, because $AF^2\; =\; BH\; \backslash times\; hB$ (CONIC SECTIONS, part iii. prop. 16, cor. 4), and $HB\; \backslash times\; hB\; =\; DH^2\; -\; DB^2$ (GEOM. part i. sect. iv. theor. 12), therefore $AF^2\; =\; DH^2\; -\; DB^2$, and $DB^2\; =\; HD^2\; -\; DG^2$. Hence it appears, that if the figure be conceived to revolve about $CA$ as an axis, so that the hyperbolic arc $AB$ may generate a hyperboloid, the triangle $DCH$ a cone, and the rectangle $DAFG$ a cylinder, any section of the first of these solids by a plane $Hh$ perpendicular to the axis, will be equal to the difference of the sections of the other two by the same plane.

Therefore the hyperboloid $BAb$ is equal to the difference between the conic frustum $FHhf$ and the cylinder $FGgf$. Let $Aa$, the transverse axis, be denoted by $p$. $Ff$ its conjugate axis by $q$, $AD$ the height of the solid by $h$, $Bb$ its base by $b$. Then, because by similar triangles, &c.

Gauging. Now $Ff^2\; =\; q^2$, and $Hh^2\; (=Bb^2\; +\; Ff^2)\; =\; b^2\; +\; q^2$, therefore putting $n$ for $.7854$, we have (by prob. 6.) the content of the conic frustum $FHhf$ equal to

from this subtract $nhq^2$, the expression for the content of the cylinder $FGgf$, and there will remain

for the content of the hyperboloid. But from the nature the hyperbola

therefore $\backslash frac\{2hq^2\}\{p\}\; =\; \backslash frac\{pb^2\}\{2(p+h)\}$; and hence the content of the hyperbola is also equal to

Now if it be considered that the quantity $nhb^2$ is the expression for the content of a cylinder whose base is $b$ and height $h$, it will appear evident, that this last formula is the same as would result from the foregoing rule.

Ex. Suppose the height of the hyperboloid to be 10, the radius of its base 12, and its transverse axis 30. What is its content?

1. Because a cylinder of the same base and altitude is $24^2\; \backslash times\; .7854\; \backslash times\; 10$, therefore, we have the proportion,

$\backslash frac\{24^2\; \backslash times\; .7854\; \backslash times\; 10\; \backslash times\; 110\}\{40\; \backslash times\; 3\; \backslash times\; 2\}\; =\; 2073.456$, the content of the solid as required.

OF GAUGING.

GAUGING treats of the measuring of casks, and other things falling under the cognizance of the officers of the excise, and it has received its name from a guage or rod used by the practitioners of the art.

From the way in which casks are constructed, they are evidently solids of no determinate geometrical figure. It is, however, usual to consider them as having one or other of the four following forms:

1. 1. The middle frustum of a spheroid.
2. 2. The middle frustum of a parabolic spindle.
3. 3. The two equal frustums of a paraboloid.
4. 4. The two equal frustums of a cone.

We have already given rules by which the content of each of these solids may be found in cubic feet, inches, &c. But as it is usual to express the contents of casks in gallons, we shall give the rules again in a form suited to that mode of estimating capacity. Observing that in each case the lineal dimensions of the cask are supposed to be taken in inches.

PROBLEM I.

To find the content of a cask of the first, or spheroidal variety.

RULE.

To the square of the head diameter add double the square of the bung diameter, and multiply the sum by the length of the cask. Then let the product be multiplied by $.0009442$, or divided by $1059.1$ for imperial gallons.

The truth of this rule may be proved thus. Put $B$ for

$FG$, the bung diameter, (see fig. next page) $H$ for $AH$ the head diameter, and $L$ for $AD$, the length of the cask, then (by prob. 14.) the content of the cask is $(2B^2\; +\; H^2)L$.

$\backslash times\; .7854$, which being divided by $277.274$ (the cubic inches in an imperial gallon) gives $(2B^2\; +\; H^2)L\; \backslash times\; .0009442$, or $(2B^2\; +\; H^2)L\; \backslash times\; \backslash frac\{1\}\{1059.098\}$ for the content in imperial gallons.

Ex. Suppose the bung and head diameters to be 32 and 24, and the length 40 inches. Required the content?

Here $(2\; \backslash times\; 32^2\; +\; 24^2)\; \backslash times\; 40\; \backslash times\; .0009442\; =\; 99.103$ imperial gallons.

PROBLEM II.

To find the content of a cask of the second, or parabolic spindle form.

RULE.

To the square of the head diameter add double that of the bung diameter, and from the sum take $\backslash frac\{2\}\{3\}$, or $\backslash frac\{1\}\{3\}$ of the square of the difference of the said diameters. Then multiply the remainder by the length, and the product multiplied, or divided by the same number as in the rule to last problem, will give the content.

For by problem 12, the content in inches is

and this formula may be otherwise expressed thus,

and hence is derived the rule, the multiplier or divisor being evidently the same as in last problem.

Ex. The dimensions of a cask being the same as in last problem; required the contents?

Ans. $(2\; \backslash times\; 32^2\; +\; 24^2\; -\; \backslash frac\{2\}\{3\}\; \backslash times\; 8^2)\; \backslash times\; 40\; \backslash times\; .0009442\; =\; 98.136$, the content in imperial gallons.

PROBLEM III.

To find the content of a cask of the third or paraboloidal variety.

RULE.

To the square of the bung diameter add the square of the head diameter, and multiply the sum by the length; then, if the product be multiplied by $.001416$, or divided by $706.2$, the result will be the content in imperial gallons.

For by problem 10, the content in inches is $\backslash frac\{1\}\{2\}(B^2\; +\; H^2)\; \backslash times\; .7854\; L$; and this expression being divided by $277.274$

gives $(B^2\; +\; H^2)\; \backslash times\; .001416\; L$, or $(B^2\; +\; H^2)\; \backslash times\; \backslash frac\{1\}\{706.2\}\; \backslash times\; L$ for the content in imperial gallons.

Ex. Suppose the dimensions of a cask, as before; required the content?

Ans. $(32^2\; +\; 24^2)\; \backslash times\; 40\; \backslash times\; .001416\; =\; 90.62$, the content in imperial gallons.

PROBLEM IV.

To find the content of a cask of the fourth or conical variety.

RULE.

To three times the square of the sum of the diameters add the square of the difference of the diameters; multiply the sum by the length; and multiply the result by $.00236$ or divide it by $4237.3$, for the content in imperial gallons.